# Replace every array element by sum of previous and next

Given an array of integers, update every element with sum of previous and next elements with following exceptions.
a) First element is replaced by sum of first and second.
b) Last element is replaced by sum of last and second last.

Examples:

Input : arr[] = { 2, 3, 4, 5, 6}
Output : 5 6 8 10 11
Explanation: We get the following array as {2+3, 2+4, 3+5, 4+6, 5+6}

Input : arr[] = { 3, 2, 1}
Output : 5 4 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A Simple Solution is to create an auxiliary array, copy contents of given array to auxiliary array. Finally traverse the auxiliary array and update given array using copied values. Time complexity of this solution is O(n), but it requires O(n) extra space.

An efficient solution can solve the problem in O(n) time and O(1) space. The idea is to keep track of previous element in loop. Add the previous element using the extra variable and the next element to get each element.

Below is the implementation of this idea.

## C++

 // C++ program to update every array element with // sum of previous and next numbers in array #include using namespace std;    void ReplaceElements(int arr[], int n) {     // Nothing to do when array size is 1     if (n <= 1)         return;        // store current value of arr[0] and update it     int prev = arr[0];     arr[0] = arr[0] + arr[1];        // Update rest of the array elements     for (int i = 1; i < n - 1; i++) {            // Store current value of next iteration         int curr = arr[i];            // Update current value using previews value         arr[i] = prev + arr[i + 1];            // Update previous value         prev = curr;     }        // Update last array element separately     arr[n - 1] = prev + arr[n - 1]; }    // Driver program int main() {     int arr[] = { 2, 3, 4, 5, 6 };     int n = sizeof(arr) / sizeof(arr[0]);        ReplaceElements(arr, n);        // Print the modified array     for (int i = 0; i < n; i++)         cout << arr[i] << " ";     return 0; }

## Java

 // Java program to update every array element with  // sum of previous and next numbers in array    import java.io.*;    class GFG {         static void ReplaceElements(int arr[], int n) {          // Nothing to do when array size is 1          if (n <= 1) {              return;          }             // store current value of arr[0] and update it          int prev = arr[0];          arr[0] = arr[0] + arr[1];             // Update rest of the array elements          for (int i = 1; i < n - 1; i++) {                 // Store current value of next iteration              int curr = arr[i];                 // Update current value using previews value              arr[i] = prev + arr[i + 1];                 // Update previous value              prev = curr;          }             // Update last array element separately          arr[n - 1] = prev + arr[n - 1];      }     // Driver program             public static void main (String[] args) {            int arr[] = {2, 3, 4, 5, 6};          int n = arr.length;          ReplaceElements(arr, n);          // Print the modified array          for (int i = 0; i < n; i++) {              System.out.print(arr[i] + " ");          }      }  }  // This code is contributed by akt_mit

## Python 3

 # Python 3 program to update every array # element with sum of previous and next  # numbers in array    def ReplaceElements(arr, n):        # Nothing to do when array size is 1     if (n <= 1):         return        # store current value of arr[0]      # and update it     prev = arr[0]     arr[0] = arr[0] + arr[1]        # Update rest of the array elements     for i in range(1, n - 1):            # Store current value of          # next iteration         curr = arr[i]            # Update current value using          # previews value         arr[i] = prev + arr[i + 1]            # Update previous value         prev = curr        # Update last array element separately     arr[n - 1] = prev + arr[n - 1]    # Driver Code if __name__ == "__main__":        arr = [ 2, 3, 4, 5, 6 ]     n = len(arr)        ReplaceElements(arr, n)        # Print the modified array     for i in range(n):         print (arr[i], end = " ")    # This code is contributed  # by ChitraNayal

## C#

 // C# program to update every array element with  // sum of previous and next numbers in array  using System; public class GFG {         static void ReplaceElements(int []arr, int n) {         // Nothing to do when array size is 1          if (n <= 1) {             return;         }            // store current value of arr[0] and update it          int prev = arr[0];         arr[0] = arr[0] + arr[1];            // Update rest of the array elements          for (int i = 1; i < n - 1; i++) {                // Store current value of next iteration              int curr = arr[i];                // Update current value using previews value              arr[i] = prev + arr[i + 1];                // Update previous value              prev = curr;         }            // Update last array element separately          arr[n - 1] = prev + arr[n - 1];     }    // Driver program      public static void Main() {         int []arr = {2, 3, 4, 5, 6};         int n = arr.Length;            ReplaceElements(arr, n);            // Print the modified array          for (int i = 0; i < n; i++) {             Console.Write(arr[i] + " ");         }     } }    // This code is contributed by Rajput-JI

## PHP



Output:

5 6 8 10 11

Time Complexity – O(N)

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